Determine the magnitude and direction of the electric field 21.7 cm directly above an isolated $33.0 \times 10^{-6} \textrm{ C}$ charge.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Electric field points radially outwards from a positive point charge. And so that means above this point charge, the electric field is gonna be pointing straight up because the electric field points in the direction of force that would be applied if we were to put a little positive test charge here. And if we did put a positive test charge here, it would be repelled by this one here, and so it will be forced upwards. The strength of the electric field is gonna be kQ over r squared. That's the formula for electric field strength around a point charge. So that's 8.988 times 10 to the nine Newton meters square per Coulomb squared. Coulomb's constant times 33 times 10 to the minus 6 Coulombs divided by .217 meters. That's our distance from this charge. Square that distance and we get 6.30 times 10 to the six Newtons per Coulomb, upwards is the electric field there.